I'm in the following situation:
Consider a centred Gaussian measure $\mu_0$ on a separable Hilbert space $X$ with covariance operator $Q \in \mathcal{L}(X)$ (positive definite, self-adjoint, trace class). Denote the Cameron-Martin space of $\mu_0$ by $E := Q^\frac12(X)$ with $\|\cdot\|_E := \|Q^{-\frac12}\cdot\|$. Let $(u_n)_{n\in\mathbb{N}}$ be a sequence in $X$ that is unbounded with respect to $\|\cdot\|_E$, but converges towards $\bar u \in E$ with respect to $\|\cdot\|_X$. Moreover, let $\varepsilon_n \to 0$ with $\varepsilon_n > 0$ for all $n \in \mathbb{N}$.
Now I want to show, that $\lim\sup_{n \to \infty} \frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(\bar u))} \le 1$, where $B_\varepsilon(u) := \{x \in X: \|x - u\|_X \le \varepsilon\}$.
Are there ideas how to prove this or can anyone recommend me according literature?
Remark: We have $\lim_{n\to\infty} \frac{\mu_0(B_{\varepsilon_n}(0))}{\mu_0(B_{\varepsilon_n}(\bar u))} = e^{\frac12\|\bar u\|_E^2}$, so we can consider $\frac{\mu_0(B_{\varepsilon_n}(u_n))}{\mu_0(B_{\varepsilon_n}(0))}$ alternatively.